Problem: Simplify the following expression: $y = \dfrac{7x^2- 18x- 9}{7x + 3}$
Answer: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(7)}{(-9)} &=& -63 \\ {a} + {b} &=& &=& {-18} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-63$ and add them together. Remember, since $-63$ is negative, one of the factors must be negative. The factors that add up to ${-18}$ will be your ${a}$ and ${b}$ When ${a}$ is ${3}$ and ${b}$ is ${-21}$ $ \begin{eqnarray} {ab} &=& ({3})({-21}) &=& -63 \\ {a} + {b} &=& {3} + {-21} &=& -18 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({7}x^2 +{3}x) + ({-21}x {-9}) $ Factor out the common factors: $ x(7x + 3) - 3(7x + 3)$ Now factor out $(7x + 3)$ $ (7x + 3)(x - 3)$ The original expression can therefore be written: $ \dfrac{(7x + 3)(x - 3)}{7x + 3}$ We are dividing by $7x + 3$ , so $7x + 3 \neq 0$ Therefore, $x \neq -\frac{3}{7}$ This leaves us with $x - 3; x \neq -\frac{3}{7}$.